Verified solution · Exam P
This Exam P sample reference tests Conditional Probability. This is a conditional probability within the subgroup of policyholders receiving the specified treatment. Of the 88 treated members, 28 belong to the low-or-normal-rate category, so the requested probability is 28/88=7/22, approximately 0.318, and choice C.
Independent solution
Setup
Conditioning on treatment restricts the denominator to treated policyholders from both heart-rate categories.
Model
Within that restricted group, the favorable count is the number from the low-or-normal category.
Compute
Reduce the count ratio and convert it to a decimal.
Answer
Approximately 31.8% of the treated subgroup belongs to the requested category.
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Original practice · fully worked
Three independent sensor readings are uniformly distributed from 0 to 1. The system reports that the largest reading exceeded 0.80. Given this report, calculate the probability that the smallest reading was below 0.20.
All three readings stay at or below 0.80 with probability 0.512, so the high-alert probability is 0.488. Inclusion-exclusion gives joint probability 0.192 for a reading below 0.20 and another above 0.80. Dividing by the alert probability gives 24/61, so choice C.
Setup
Let H be the event that the maximum exceeds 0.80 and L the event that the minimum is below 0.20.
Model
Use inclusion-exclusion. The complement of H keeps every reading at most 0.80, while the complement of L keeps every reading at least 0.20.
Compute
Calculate the joint event and divide by the reporting probability.
Answer
The conditional probability is approximately 0.393.
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ActuaryProof. “SOA Exam P Sample Question #377 — Verified Solution.” Verified 2026-07-31.
Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.
